theorem Th85:
  B\orC\and\notC\impB in F
  proof
    (C\and\notC\impB) in F & B\impB in F by Th34,Def38; then
    B\orC\and\notC\impB\orB in F & B\orB\impB in F by Th59,Th52;
    hence B\orC\and\notC\impB in F by Th45;
  end;
